Difference between multiple normal populations

Suppose I know that the diastolic blood pressure of my 30 patients is normally distributed with the same mean and standard deviation. Then the odds of one patient having a diastolic blood pressure is x higher than the others is the probability that the observed value of a normally distributed variable, with mean=0 and sd=sqrt(variance * 30), equals x.
Am I right?

It sounds like you have N=30 patients, presumed to be selected randomly (sampled) from a population which has BP ~ N(m,s).

Do you want to know how likely that the extremal BP from the patients differs from the sample mean by a given amount x? (Which is literally what you asked.)

Or, do you want to know how likely it is that there are two patients in your sample with BP values that differ by a particular value?

Note - even that is unclear - if BP is a continuous random variable, then the probability two patients' BP differs by exactly x will be zero.

Note - the difference between two independant random normal variables would, itself, be distributed randomly with a mean of zero and a standard deviation of root-two times. Which you seem to have noticed - but for some reason you seem to be trying to subtract somehow between all your patients.

OK - from that clarification:
I'll state the question I'm answering and then guide you to the method of answering it.

If bi is the BP of the ith patient in the sample of N=30 selected out of a large population whose BP is distributed as a normal with mean m and standard deviation s, then,

you want to know the probability that at least one bi falls outside the range R of all possible blood pressures such that m-x < b < m+x .

Your starting point is to figure out the probability that all 30 patients BP's lie inside R - which you can get from the individual probability and that patient BPs are usually independent.

I think the core understanding you need here is that the probability that a particular patient's BP lies in the range x<b<y is given by the area under the graph:[tex]P(x<b_i<y)=\frac{1}{s\sqrt{2\pi}}\int_x^y e^{\frac{(b-m)^2}{2s^2}}db[/tex]

If you are asking for the probability that the largest of 30 normal random variables differs from the mean by such-and-such, then you are dealing with a non-normal distribution. To understand this, let F(x) = normal cdf with your given mean and standard deviation. For a sample of size n = 30, the cdf of the largest of these (i.e., the maximum) is [itex] G(x) = F(x)^{30}, [/itex] so the probability that the max exceeds y is [itex] 1-F(y)^{30}.[/itex] There are no simple formulas for this; you need to do a numerical computation. We do not even have any nice results for the expected value or variance of the maximum.